BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

last SG · 2010-07-26T23:40:35+00:00

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why iodine is a solid at rtp but hydrogen iodide is a gas?

Because iodine (although non-polar) has very polarizable electron charge clouds (due to it's large no. of electrons and large atomic radius) and hence relatively stronger instantaneous dipole - induced dipole van der Waals (ie. London dispersion forces), in comparison to hydrogen iodide, which may be polar, but still has weaker permanent dipole - permanent dipole van der Waals (ie. Keesom forces) because it has a lot less electrons and H is much smaller, and accordingly it's molecular electron charge clouds are hence much less polarizable.

Polarity is therefore outweighed by no. of electrons, size of atom/molecule and hence polarizability of electron charge clouds. Permanent dipole - permanent dipole van der Waals (ie. Keesom forces) is only stronger than instantaneous dipole - induced dipole van der Waals (ie. London dispersion forces) in cases where the no. of electrons, size of atom/molecule, and hence hence polarizability of electron charge clouds, are comparable or similar.

draw an isomer of octane which contains chiral carbons but is optically inactive. is that possible?

Yes, such molecules are called meso compounds (assuming we're not talking about a racemic mixture of both enantiomers), which contain one or more chiral carbons, but are optically inactive due to either an internal plane of symmetry, and/or an internal center of inversion.

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KickMe asked :

When dry chlorine is passed over heated aluminium foil in a hard glass tube, a vapour is produced which condenses to a yellow-white solid on the cooler parts of the tube. At low temperatures, the vapour has the empirical formula. AlCl3, and an Mr of 267. I understand that a dimer was formed. However, I do not know what the yellow-white solid is. Can you tell me?

The yellow-white solid is simply solid aluminium chloride, which has a (sheet-like layered cubic) covalent polymeric structure or ionic lattice structure (both description adjectives are acceptable and correctly describe the same structure, because there is both ionic and covalent character present in the bonds) in the solid state, as opposed to a simple covalent molecular structure in the liquid state (molecular dimer) or in the gaseous states (dimers at lower temperatures, monomers at higher temperatures).

The H2 syllabus only requires you to be able to draw the structures in the liquid and gaseous states (ie. the molecular dimer and monomer), so it's unlikely you'll be asked to draw the structure of AlCl3 in the solid state, but you should at least be aware that in the solid state it has a ionic or polymeric lattice structure, and not simple molecular.

As your chemistry basics would have told you, simple molecular compounds are usually liquid or gaseous, not solid.

Check out an image of AlCl3 in the solid state here on Rod Beavon's webpage :
http://www.rod.beavon.clara.net/aluminiu.htm

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2011 'A' level exam qn :

Is PCl5 a simple molecular compound? If so, why is it solid at rtp? Draw it's structure in the solid state and describe the geometries involved.

Answer :

PCl5 exists as a simple molecular compound in the liquid state, but exists as ionic PCl4+ PCl6- in the solid state. The ionic geometries are tetrahedral and octahedral respectively for cation and anion.

The P-Cl bond dissociation enthalpy is small, compared to the large amount of heat energy that can be evolved from forming the stronger ionic bonds. Hence formation of solid PCl5 is favoured under standard conditions.

Simple molecular compounds (eg. water, methane, etc) are almost always liquid or gaseous at rtp, with the notable exception of iodine, for which the large no. of electrons and the large atomic radius, results in highly polarizable electron charge clouds, resulting in extensive and relatively strong instantaneous dipole - induced dipole van der Waals (ie. London dispersion forces), and hence iodine is a simple covalent molecular solid at room temperature.

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kickme asked :

I was asked to indicate clear the type of hybridisation on each carbon atom of a hydrocarbon. I know the hybridisation state of the carbon atoms. However, I am rather unsure of how I should present the hybridisation state.

Is it alright for me to draw the entire structure of the hydrocarbon out and just note the hybridisation state at the side of the carbon atom by writing sp, sp2 or sp3? Or is there a proper chemistry way of presenting such an answer?

That's fine. You can also circle the individual C atoms and use long arrows to link it to your written answers "sp3" etc at the side or below the structure if you like.

Bottomline, as long as it's clear and unambiguous communication, it's good Chemistry, whether at 'A' levels or at any level (including professional career Chemists).

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KickMe asked :

Phosgene dissociates to form CO and Cl2

COCl2 (g) <----> CO (g) + Cl2 (g)

The degree of dissociation, a, of phosgene is different at different temperature. 9.9g of phosgene is allowed to dissociate in a 10dm3 vessel and the equilibrium mixture attains a pressure of 6.55x10^4 Pa at 633K.

a) Calculate the degree of dissociation, alpha (a), of phosgene under the above conditions.

PV=nRT
P=nRT/V=0.1(8.31)(633)/0.01=5.26x10^4

COCl2 <------> CO + Cl2
Initial 5.26x10^4 0 0
Change -a(5.26x10^4) +a(5.26x10^4) +a(5.26x10^4)
Eqm. (1-a)(5.26x10^4) a(5.26x10^4) a(5.26x10^4)

(1-a)(5.26x10^4)+a(5.26x10^4)+a(5.26x10^4)=6.55x10^4
(1+a)(5.26x10^4)=6.55x10^4
a=0.245

The next part then asks calculate the equilibrium concentration of COCl2, CO and Cl2.

I used PV=nRT
n=PV/(RT)

since concentration= n/v = P/(RT)

is it appropriate to do so?

Yes, your working is correct.

As for concentration, unless otherwise specified, it refers to molarity, which is moles / dm3. Hence, your V in your PV=nRT should be in dm3, if you intend n/V to be molarity.

When V is in m3 and P is in Pa (ie. 1 atm = 1.01325 x 10^5 Pa), R = 8.314.
When V is in dm3 and P is in atm (ie. 1 atm = 1.01325 x 10^5 Pa), R = 0.08206.

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kickme posted :

Now that, I know the yellow white solid is aluminium chloride, AlCl3. The next part of the question asks:

When a few drops of water are added to the solid, streamy white fumes are evolved and a white solid remains, which is insoluble in water.

Can I have some clue as to what reaction occurs when water is added to solid aluminium chloride?

First of all, take note that the reaction and equations commonly tested at 'A' levels involve adding excess water to generate an acidic solution of AlCl3(aq), which is really [Al(H2O)5(OH)]2+(aq) + H3O+(aq) + 3Cl-(aq).

However, this question is asking what happens when limiting amounts of water are added to excess AlCl3(s).

Even though the initial reactant AlCl3 may be regarded as covalent, but because the final product is more ionic than covalent, and has the empirical formula Al2O3, it'll be more easier for you to follow mathematically if I illustrate the reaction mechanism using two units of AlCl3 rather than one.

Six (per two units of AlCl3) water nucleophiles / ligands / Lewis bases attack (ie. donate dative bonds to) the (high charge density and thus a strong) Lewis acid Al3+ ion of AlCl3 (empirical formula), thereby eliminating six (per two units of AlCl3) Cl- ions which instantaneously functions as a Lewis / Bonsted base to abstract the six (per two units of AlCl3) protons from the positive formal charged (O atoms of the) six water nucleophiles which attacked earlier, consequently generating the steamy fumes of six HCl(g) molecules and the intermediate Al2(OH)6 which readily dehydrates (due to the high charge density or 'electrophilicity' of the Al3+ ion making it a strong Lewis acid, half of the OH- ligands / nucleophiles / Lewis bases present readily donate a second dative bond and thus loses another proton; furthermore the reaction is exothermic and the high temperature provides additional activation energy *and* favours the positive increase in entropy as predicted by Gibbs free energy during the dehydration process) to generate the insoluble white solid Al2O3 (empirical formula) and three water molecules are simultaneously eliminated (accounting : these three H2O molecules came from the elimination of the 3 OH- ligands/nucleophiles combining with the 3 protons eliminated, when the other 3 OH- ligands/nucleophiles donated a 2nd dative bond to the Al3+ ion).

But because the ions involved are O2- and Al3+, and since the electronegativity difference between oxygen and aluminium is sufficiently large to qualify as predominantly ionic, the so-called 'dative covalent bonds' donated to the Al3+ during the mechanism are now to be regarded as ionic bonds, in the final ionic product Al2O3.

Hydolysis (ie. nucleophilic / Lewis base attack of H2O) :
2 AlCl3 + 6 H2O → 6 HCl + 2Al(OH)3 or Al2(OH)6

Dehydration of Al(OH)3 :
2 Al(OH)3 or Al2(OH)6 → Al2O3 + 3 H2O

Overall :
2 AlCl3(s) + 6 H2O(l) → Al2O3(s) + 6 HCl(g) + 3 H2O(l)

Cancelling the three H2O molecules on both sides of the equation :
2 AlCl3(s) + 3 H2O(l) → Al2O3(s) + 6 HCl(g)

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Kickme asked :

I have another doubt pertaining to the 'A' level question asked ealier: 2011 'A' level exam qn :

Is PCl5 a simple molecular compound? If so, why is it solid at rtp? Draw it's structure in the solid state and describe the geometries involved.

In your answer, you stated: The P-Cl bond heterolytic dissociation enthalpy is small, compared to the large amount of heat energy that can be evolved from forming the stronger ionic bonds. Hence formation of solid PCl5 is favoured under standard conditions.

I understand that the P-Cl bond dissociation enthalpy is referring to the formation of simple molecular PCl5. But I dont understand how the formation of simple molecular PCl5 requires the bond dissociation of P-Cl.

Unless, simple molecular PCl5 also exists as PCl4+ and PCl6- ions initially, and PCl6- loses a Cl- to donate to PCl4+, thus forming 2 PCl5 molecules.

If we begin from the molecular PCl5, then the P-Cl heterolytic (not homolytic) bond dissociation generates the ionic species PCl4+ and PCl6-, as temperature decreases and PCl5 solidifies.

As temperature increases, because molecular PCl5 is able to possess more kinetic energy as liquid and gaseous simple molecules, compared to solid ionic PCl5 (whose atoms can only vibrate about fixed positions), hence the P-Cl heterolytic (not homolytic) bond dissociation now generates the molecular PCl5 from the ionic species.

In other words, whether you're generating the ionic from the molecular, or the molecular from the ionic, you'll need to heterolytically dissociate a P-Cl bond (and of course, form one as well). The molecular and ionic species exist in equilibrium with each other, it's only a matter of who predominates in which temperatures, pressures and solvents.

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Memo wrote to Kickme :

I don't know if you have figured this out. but if this ever come out in exam i would write something like this...

As the total number of electrons per molecule increases from (HCl->HBr->HI), more electrons can be polarised. Thus the strength of the Van der Waals' forces due to induced dipole - induced dipole increases.

However, the polarity of the H-X bond decreases from (H-Cl->H-Br-H-I). This results in weaker Van der Waals' forces due to permanent dipole - permanent dipole attraction.

The increase in strength of the Van der Waals' forces due to induced dipole - induced dipole attractions outweighs the decrease in strength of permanent dipole - permanent dipole attractions.

Hence, the boiling point of (HCl > HBr > HI).

Kickme wrote :

Shouldn't it be instead: the induced dipole-induced dipole van der Waals between HI molecules are stronger than the permanent dipole-permanent dipole van der Waals between HCl molecules.

Since the greater electronegativity of Cl over I already shows that HCl has got stronger permanent dipole-permanent dipole interactions over HI.

I replied them :

Ok, here's the thing :

Both of you guys (Memo and Kickme) are stuck in the thinking that the strength of the permanent dipole - permanent dipole van der Waals are determined solely by the electronegativity difference, while the strength of instantaneous dipole - induced dipole van der Waals are determined solely by the polarizability of electron charge clouds.

This is what is commonly taught in most JCs, and is perfectly acceptable by Cambridge.

But while acceptable (if you feel comfortable with this concept, go ahead and write Memo's answer, it's perfectly acceptable). it is not completely accurate. If you reread my earlier post, there's another way of looking at it (equally acceptable by Cambridge), which is :

The strength of any of the three types of van der Waals (Keesom, Debye, London dispersion forces), is first and foremost determined by polarizability of electron charge clouds, and next by polarity of electronegativity differences.

If you follow this concept, you'll still end up with the same correct conclusion. Use my concept, or use Memo's. Either way, you get the same conclusion, and full marks from Cambridge.

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The products obtained by cracking an alkane, X, are methane, ethene and propene.
The mole fraction of ethene in the products is 0.5.

What is the identity of X?

A C6H14 B C8H18 C C9H20 D C11H24

Start by drawing these 3 smaller hydrocarbon products and piece them together like a jigsaw puzzle. Based on the fact that the longest hydrocarbon obtained is only 3 carbons, obviously the original large alkane must be branched. Based on the fact that 50% of the products are ethene, that means two of the four branches must be ethyl groups. The third branch must be the propyl group. The intersection of the branches, itself generates methane.

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Kai asked :

1. a 0.10M solution of Pb(NO3)2 is added enough HF(g) to make [HF]= 0.10M. (Ksp of PbF2= 2.7e-8 , Ka of HF= 6.6e-4)
(i) Does PbF2 precipitate from this solution?
(ii)What is the minimum pH at which PbF2 precipitates?

2. Calculate the molar solubility of AgCN in a buffer solution with a pH of 3.00. (Ksp of AgCN=1.2e-16 Ka of HCN= 6.2e-10 )

The ans. to these two problems are: 1: (i) yes (ii) 0.90 2: 1.4e-5 mole/L

SnowYak replied him :

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Atomos asked :

Hi all,

I have a question on the electronic configuration of particles, especially when it uses the idea of the stability of 3d^5 4s^1

Some scenarios might be more hypothetical than feasible.

Case 1

What is the final configuration when a particle with initial configuration [Ar] 3d^3 4s^2 gains an electron?

[Ar] 3d^4 4s^2 or [Ar] 3d^5 4s^1?

Case 2

What is the final configuration when a particle with initial configuration [Ar] 3d^6 4s^2 loses two electrons?

[Ar] 3d^6 or [Ar] 3d^5 4s^1?

Too many conflicting approaches on the net, hope to clear things up here.

Thanks in advance!

Case 1 :

You're right, this is more hypothetical than feasible, because vanadium is a metal that is only interested in losing electrons to be oxidized to cationic form, not gaining electrons to be reduced to anionic form.

The short answer is : both configurations are possible, depending on other factors (that you needn't concern yourself with at 'A' levels).

However, the more important lesson here is, the exam-smart candidate will :

If this was a P1 qn, then consider both possible, but (if both options are present, then) assume the more direct answer, ie. 3d4 4s2.

If this was a P2 or P3 qn, then intelligently (ie. with explanation) write both configurations in your answer :

"When a particle with initial configuration [Ar] 3d^3 4s^2 gains an electron, it first attains an electron configuration of [Ar] 3d^4 4s^2, which may (or may not) subsequently rearrange itself (as observed in the example of chromium) to have an electron configuration of [Ar] 3d^5 4s^1, by transferring an electron from its 4s orbital, to one of its 3d orbitals, to achieve a more stable half-filled 3d subshell."

Case 2 :

This is pretty straightforward. No rearrangement takes place (as far as the electron configuration of the Fe2+ ion is concerned, when asked at 'A' levels) because Fe2+ is ionic and readily stabilizes itself in other ways (ie. as an alternative to rearranging its electron configuration) such as by forming ionic bonds, and/or coordination complexes (the electrons-in-boxes diagram will include both its own electrons, as well as those of the coordinate dative bonds donated by the ligands).

Bottomline :

At 'A' levels, when removing electrons from d-block elements, remember to just kick away electrons from the 4s orbital (which is further away from the nucleus, thus easier to remove), before removing any electrons from its 3d orbital (which is nearer to the nucleus, thus more difficult to remove).

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Explain why despite nitrogen and phosphorus both being in Group V, nitrogen forms N2 molecules while phosphorus forms P4 molecules.

Ans : see my website ( google "BedokFunland JC" )

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kickme asked :

A and B are isomers and they have the molecular formula C3H5Cl. B is able to decolourise orange aqueous bromine to form C, while A does not decolourise bromine. When ethanolic KCN is heated with A and B, then subsequently further heated with dilute nitric acid, only A forms a product D. D is able to react with sodium carbonate to give of a colourless gas and an ion, E. When A is heated with limited amount alcoholic ammonia, a white solid F is formed.

Draw all the full displayed formula of A, B, C, D, E and F. Write the balaced equation for all the reaction that has taken place.

A: chlorocyclopropane

B: 3-chloropropene

C: 1-bromo,3-chloropropa-2-ol

D: cyclopropanoic acid

E: I am not sure, but is the ion from the earlier part when A reacts with dilute HNO3 to form NH4+ ion?

F: Is the reaction through polyalkylation? But i do not know how to write the balanced equation for a cycloalkane!

The balanced equation part should be fine except for F. Please help to check if the compound A to F identified is correct. Thank you!

Good attempt, with a couple of errors. Here are the correct answers.

A : chlorocyclopropane

B : 1-chloroprop-1-ene or 2-chloroprop-1-ene (the halogen atom must be bonded to a sp2 hybridized C atom, allowing resonance delocalization of a lone pair on the halogen atom to form a pi bond with the C atom, thus giving rise to C-X having partial double bond character, as evidenced by the observed resistance to nucleophilic substitution by the CN- nucleophile)

C : (based on either species B above)
either
1-bromo-1-chloropropan-2-ol (major product) or 2-bromo-1-chloro-propan-1-ol (minor product)
or
1-bromo-2-chloropropan-2-ol (major product) or 2-bromo-2-chloro-propan-1-ol (minor product)

D : cyclopropylmethanoic acid

E : sodium cyclopropylmethanoate

F : tetra(cyclopropyl) ammonium chloride (with byproduct of HCl(alc) <---> HCl(g) )

About writing the balanced equation to generate F, you can either write the 4 individual equations (better yet, draw the repeated nucleophilic subsitution mechanism), or a combined overall equation (to generate 3 HCl and 1 RRRRN+Cl-)

Cambridge will accept either, since they didn't specify which was required.

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KickMe then replied :

thank you very much.

just some questions, since the question state that E is an ion, as Nish said, the answer should be cyclopropylmethanoate ion, right?

But since sodium is present, wouldn't sodium react with the ion to form the salt sodium cyclopropylmethanoate?

Another question is about the naming, is cyclopropylmethanoic acid=cyclopropanoic acid?

Cambridge will usually ask you to identify unknown compounds such as E, as a full ionic compound (ie. a metal carboxylate, in this case, sodium cyclopropylmethanoate), rather just as one of the ions present. But yes, if the prelim exam question specifies E as an ion, they're indeed asking for "cyclopropylmethanoate".

In aqueous or alcoholic solvent, sodium compounds are soluble, meaning that the Na+ cation and the cyclopropylmethanoate anion, are both ion - permanent dipoled (or for the anion, hydrogen bonded) to the water or alcohol solvent, rather than ionically 'stuck' to each other, which would be the case if the solvent was aprotic non-polar (eg. CCl4, methylbenzene, etc).

No, as Nish pointed out already, the name "cyclopropanoic" is incorrect because it implies only 3 Cs are present. The correct name "cyclopropylmethanoic acid" implies there are 3 Cs + 1 C = 4 Cs present in total, which is indeed the case here.

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To add on another question is: I didn't quite understand why the halogen must be bonded to the sp2 carbon for B.

You said: "B : 1-chloroprop-1-ene or 2-chloroprop-1-ene (the halogen atom must be bonded to a sp2 hybridized C atom, allowing resonance delocalization of a lone pair on the halogen atom to form a pi bond with the C atom, thus giving rise to C-X having partial double bond character, as evidenced by the observed resistance to nucleophilic substitution by the CN- nucleophile)"

It is the same concept or explanation with "why doesn't chlorobenzene undergo hydrolysis? in contrast to chloroalkanes".

Most JCs give the orbital overlap explanation, while I personally prefer the resonance delocalization explanation. Cambridge will accept both, as long as you specify that the end-result is that the C-X bond has partial double bond character and is difficult to cleave, and therefore the halobenzene or haloalkene, is resistant to nuclophilic substitution.

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BlackToast asked :

Hi ultima!
regarding the very first post in this thread, Q1, i din't get your explanation so i went to clarify with my teacher. he said i'm supposed to use enthalpy c of soln = enthalpy c of hyd - lattice energy. BUT he din't give me the full explanation, asked me to go home and think. -_-"

By Hess Law,
Solution enthalpy = (endothermic) Lattice Dissociation enthalpy + (exothermic) Hydration enthalpy.

Your school teacher is using the exothermic version, ie. Lattice Formation enthalpy, in his formula, which will still give you the same correct mathematical answer, but makes less sense, in the context of Hess Law, and therefore I would advise you to use my formula, instead of your school teacher's.

For a full explanation on why Grp II hydroxides become more soluble down the group, but why Grp II carbonates become less soluble from Be to Sr, but mysteriously becomes more soluble again from Sr to Ba, read Jim Clarke's website (Jim Clarke's and Rod Beavon's websites, are the top two most popular 'A' level Chem websites on the internet, you should make the most use of them) :
http://www.chemguide.co.uk/inorganic/group2/problems.html

Even though solubility trends for Group II hydroxides versus carbonates & sulfates, are not required by the H2 syllabus to be memorized, Cambridge can still ask you deduce-and-explain questions on this topic. Students aiming for a distinction, should definitely study the material (Solubility of Group II hydroxides versus Carbonates) on Jim Clarke's website.

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In explaining why C-X bond is cleaved to produce X radicals and R radicals, instead of cleaving the C-H bond to produce H radicals; besides saying the H radical is much less stable, can i also say because the C-H bond is very strong?

Yes, you can quote relevant data booklet values and say the C-H bond is stronger than the C-X bond, and therefore homolytic cleavage of the C-X bond is energetically favoured over the homolytic cleavage of the C-H bond.

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So u mean if they say the temperature is cold, and there is an alchohol attached to the side chain of a benzene ring, then the alchohol will get oxidised only?

Yes, this is correct. Usually Cambridge and Prelim papers will specify "heat under reflux with KMnO4", and student is expected to give "benzoic acid" as the product. But a potential trick question, is to "heat under reflux with K2Cr2O7", in which case you only get oxidation of the alcohol (and not the entire side chain), but there will be a % of students who don't read carefully, and wrongly give "benzoic acid" as the answer.

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